Physics help needed ASAP

[Linas]

I am trying to do my physics homework, and its too late to see a tutor. Help if you can.

Running with an initial velocity of 11 m/s, a horse has an average acceleration of -1.79 m/s^2.

How long does it take for the horse to decrease its velocity to 7.0 m/s?

[N8116B]

I take from the way you ask the question you would want me to use simple algebra to solve.

The idea is simple you are traveling at 11 m/s and you want to slow down to 7 m/s
acceleration is a distance divided by a time squared. Therefore if you multiple an acceleration by a unit of time you get a velocity. When you learn calculus later this is much easier and you'll realize that acceleration is the fist derivative of velocity.

So basically at has to equal -4m/s

V1 = 11 m/s
V2 = 7 m/s
a = -1.79 m/s²
t = time it takes to decelerate to V2

V1 + at = V2

11 + (-1.79)t = 7
-1.79t = 7-11
-1.79t = -4
t = -4/-1.79
t = 2.23 seconds

[Linas]

amazing. thanks. This is my first time with physics, and i really need to see some help with it.

[N8116B]

Originally Posted by Linas

amazing. thanks. This is my first time with physics, and i really need to see some help with it.

A helpful thing is to keep track of units.

You are dealing in m/s

But acceleration is in m/s²

to get if you multiple acceleration by an 's' it looks like this:

m/s² * s

or ms/s² which is simplified to m/s as s/s is equal to 1

Can you see that?
It helps to always look at units in these types of problems.

[Fontaine]

good ole highschool physics

for future reference you going to use this one a lot:

d = Vit + 1/2 at^2

not a hard formula...but i mean a lot

[PSPookie]

Originally Posted by Fontaine

good ole highschool physics

for future reference you going to use this one a lot:

d = Vit + 1/2 at^2

not a hard formula...but i mean a lot

Yup, that's the basic displacement equation. You should also get familiar with its 1st and 2nd derivatives:

Velocity:
V(t)= Vi + at

Acceleration:
a(t)= a

If you were to graph dislacement vs. time, the velocity at any given point would be equal to the slope of the line tanget to the displacement curve at that point. A similar relationshp exists between velocity and acceleration.

Don't let the alphabet soup confuse you -- this stuff makes sense, but you have to learn the language.

+1 to the unit tracking suggestion

Drawing diagrams also helps.

[Atoz]

Yep... if you remember the "Big Three" Laws of Kinematic Motion:

V_f = V_o + at

V_f² = V_o² + 2as

s = V_ot + 1/2at²


(s = distance traveled)


You'll be OK. They got me through 1st semester physics 25 years ago.

[Steve]

How fast a horse decelerates is largly dependant on the mood of the horse. Isn't it?

[MakiSupaStar]

I just learned something. I need to go lay down.

[Techmonkey]

http://en.wikipedia.org/wiki/SUVAT_equations

exactly the same as what's been mentioned, but I find it easier to remember them that way

[PSPookie]

Originally Posted by Atoz

Yep... if you remember the "Big Three" Laws of Kinematic Motion:

V_f = V_o + at

V_f² = V_o² + 2as

s = V_ot + 1/2at²


(s = distance traveled)


You'll be OK. They got me through 1st semester physics 25 years ago.

Whoa! Look at Atoz -- gettin' fancy with the sub/superscripts

[Atoz]

Originally Posted by PSPookie

Whoa! Look at Atoz -- gettin' fancy with the sub/superscripts

I'm a scientist... I can't help myself.

[PSPookie]

Oh and one more immutable law of physics:
Your coolness is inversely proportional to the number of days it takes that cab to get to my house